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\pub{2009}{1}{3}{1}
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\topic{Lecture 3 \\Differential Calculus-I\\ \scriptsize Leibnitz's Theorem (17 Sep 2009)}
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\subsection*{Find $y_n$ at $x=0$}
\begin{quote}
If $u$ and $v$ are two functions of $x$, then
\[D^n(uv) = ^nC_0 D^nu.v + ^nC_1 D^{n-1}u.Dv+^{n-2}C_2 D^{n-2}u.D^2v+ ... + u.^nC_n D^nv\]
\end{quote}
\subsection*{Working Rule}
\begin{enumerate}
	\item Arrange function in form $y=f(x)$.
	\item Find $y_1$, then $y_2$.
	\item Differentiaite both side $n$-times using Leibnitz Theorem.
	\item Put $x$ in required equations. There may be two cases: (i) $n$ is odd or (ii) $n$ is even. Take care.
\end{enumerate}
\begin{example}
If $y=(sin^{-1}x)^2$, prove that 

$y_n(0)=0$ for $n$ odd and $y_n(0)=2.2^2.4^2.6^2 ... (n-2)^2;n \neq 2$ for $n$ even. 
\end{example}

Given $y=(sin^{-1}x)^2$
Differentiating with respect to $x$
\[
\begin{array}{rcl}
y_1 &=& \frac{2 \sin^{1}x}{\sqrt{1-x^2}}\\
 &=& \\
 &=& \\
 (1-x^2)y_1^2&=&4y
\end{array}
\]
Differentiate above equation with respect to $x$, we get
\[
\begin{array}{rcl}
 &=& \\
 (1-x^2)y_2 - xy_1 -2 &=&0
\end{array}
\]
Differentiate $n$ times above equation with respect to $x$, we get
~\\
~\\
~\\
\[(1-x^2)y_{n+2} - (2n+1)xy_{n+1} -n^2y_n &=&0\]
Now, on putting $x=0$, we get
\[
\begin{array}{rcl}
 y(0)&=&0 \\
 y_1(0)&=&0 \\
 y_2(0)&=&2 \\
\end{array}
\]
and 
\[y_{n+2}(0)=n^2y_n(0)\]

Case: When $n$ is odd, Put $n=1,3,5 ... $, we get
\[
\begin{array}{rcl}
 y_3(0)&=&1^2.y_1(0)=0 \\
 y_5(0)&=&3^2.y_3(0)=0 \\
 Thus,~~~y_7(0)&=& y_9(0) = y_{11}(0) = ... =0 
\end{array}
\]

Case: When $n$ is even, Put $n=2,4,6 ... $, we get
\[
\begin{array}{rcl}
 y_4(0)&=&2^2.y_2(0)=2^2.2 \\
 y_6(0)&=&4^2.y_4(0)=4^2.2^2.2 \\
\end{array}
\]
Hence, in general, $y_n(0)=2.2^2.4^2 ... (n-2)^2$. for $n$ is even and $n\neq2$. (Think! why?)
\section*{Problems}
\begin{enumerate}

\item   If $y=  sin^{-1} x$, prove that $(1-x^2) y_{n+2} - (2n+1) x y_{n+1}-n^2 y_n = 0$ Also find the value of $y_n$ when $x = 0$.

\item   If$y=\frac{\sin ^{-1} x}{\sqrt{1-x^{2} } } $, prove that $y_n = (n-1)^2 y_{n-2}$ for $x = 0$.

\item   If $y =[x+\sqrt{1+x^{2} } ]^{m} $, find $y_n(0)$.

\item   Find $y_n (0)$ when $y = \log(x+\sqrt{1+x^{2}})$.

\item   If $y = [\log(x+\sqrt{1+x^{2} } )]^2$, find all the derivatives of $y$ w.r.t. $x$, when $x = 0$. 

\item  If $y = (sinh^{-1}x)^2$, prove that $(1+x^2) y_{n+2} + (2n+1) x y_{n+1} + n^2 y_n = 0$. Hence find $y_n$ when $x = 0$.

\item   If $y = e^{a\sin ^{-1} x} $, prove that $(1-x^2) y_{n+2} - (2n+1) xy_{n+1} - (n^2+a^2) y_n = 0$. Deduce that $\mathop{Lt}\limits_{x\to 0} $$\frac{y_{n+2} }{y_{n} }  = n^2 + a^2$. Hence find $y_n (0)$.

\item   If $y = \sin (m \sin^{-1}x)$, find $y_n(0)$.

\item   Prove that the value when $x = 0$ of $D^n(tan^{-1}x)$ is $0,(n-1)!$ Or $-(n-1)!$  according as $n$ is of the form $2p, 4p + 1$ or $4p + 3$ respectively.

\item   If $y = \log (x+\sqrt{x^{2} +a^{2} })$, prove that $(a^2 + x^2) y_2 + xy_1 = 0$. Differentiate this equation $n$ times and prove that $\mathop{\lim }\limits_{x\to 0} \frac{y_{n+2} }{y_{n} } =-\frac{n^{2} }{a^{2} } $.

\item   If $\log y = tan^{-1}x$, show that \[(1+ x^2) y_{n + 2} + \{2(n + 1) x - 1\} y_{n + 1} + n (n+1) y_n = 0\] hence find $y_3, y_4$ and $y_5$ at $x = 0$.

\item   If $y = e^{m \cos ^{-1} x}$, show that $(1- x^2) y_{n + 2}- (2n + 1) x y_{n + 1} - (n^2 + m^2) y_n = 0$ and calculate $y_n(0)$.
\item   If $y = \tan^{-1}x$, prove that $(1+ x^2) y_{n + 1} + 2nxy_n + n (n-1) y_{n - 1} = 0$. Hence determine the values of all the derivatives of $y$ with respect to $x$ when $x =0$. \\\textbf{Ans}: When $n$ is even, $y_n(0) = 0$. When $n$ is odd, $y_n(0) = (-1)(n-1)/2(n-1)! $ ???
\end{enumerate}

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